68 lines
1.8 KiB
Java
68 lines
1.8 KiB
Java
//<p>给你一个整数 <code>n</code> ，按字典序返回范围 <code>[1, n]</code> 内所有整数。</p>
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//
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//<p>你必须设计一个时间复杂度为 <code>O(n)</code> 且使用 <code>O(1)</code> 额外空间的算法。</p>
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//
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//<p> </p>
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//
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//<p><strong>示例 1：</strong></p>
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//
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//<pre>
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//<strong>输入：</strong>n = 13
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//<strong>输出：</strong>[1,10,11,12,13,2,3,4,5,6,7,8,9]
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//</pre>
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//
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//<p><strong>示例 2：</strong></p>
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//
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//<pre>
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//<strong>输入：</strong>n = 2
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//<strong>输出：</strong>[1,2]
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//</pre>
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//
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//<p> </p>
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//
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//<p><strong>提示：</strong></p>
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//
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//<ul>
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// <li><code>1 <= n <= 5 * 10<sup>4</sup></code></li>
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//</ul>
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//<div><div>Related Topics</div><div><li>深度优先搜索</li><li>字典树</li></div></div><br><div><li>👍 277</li><li>👎 0</li></div>
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package leetcode.editor.cn;
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import java.util.ArrayList;
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import java.util.List;
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// 386:字典序排数
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public class LexicographicalNumbers {
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public static void main(String[] args) {
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Solution solution = new LexicographicalNumbers().new Solution();
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// TO TEST
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solution.lexicalOrder(13);
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}
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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public List<Integer> lexicalOrder(int n) {
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dfs(n, 1);
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return list;
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}
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List<Integer> list = new ArrayList<>();
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private void dfs(int n, int num) {
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if (num > n) {
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return;
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}
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list.add(num);
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dfs(n, num * 10);
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if (num % 10 == 0 || num == 1) {
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int max = (num / 10 + 1) * 10;
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for (int i = num + 1; i < max; i++) {
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dfs(n, i);
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}
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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